This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1880 Excerpt: ... triangle. Hence the required chance is . 6098. (By Prof. Cochez.)--Trouver dans le plan de trois cercles un point tel que les tangentes menees de ce point aux troix cercles soient entre elles comme trois droites donnees. Solution by J.'A. Kealy, M.A.; Rev. J. L. Kitchin, M.A.; and others. Let f(x, y) = 0 and f, (x, y) be two of the circles; then f(x, y) = lc. p (x, y) is the locus of a point from which the tangents to the circles are in a given ratio, and this locus must evidently be some circle. Hence we have the following geometrical construction: --Draw the four common tangents, or at least three of them, to two of the circles; divide them in the given ratio, and draw a circle through the points of division; do the same with another pair; then the intersections of these two circles will give the points required. 5538. (By Professor Townsend, F.R.S.)--A heavy flexible chain, of uniform thickness, being supposed to occupy longitudinally just half the interior of a slender circular tube, of uniform roughness, bounded radially by coaxal horizontal cylinders, and laterally by parallel vertical planes; determine, given all particulars, its two extreme positions of equilibrium under the action of gravity. Solution by the Proposer. Denoting by a the radius of the tube, by m the mass per unit of length of the chain, by 0 the angle between the radius at any point P of its length and the horizontal radius of the tube through whose extremity it does not pass, by T and E the tension of the chain and the reaction of the tube at P, and by p and t the coefficient and angle of the friction arising from the uniform roughness of the tube; then, since, by tangential and normal resolution, in the usual manner, ----/-E = amycos6, and T---R =--amaind; and since, ...

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